Luminiferous Ether

Discussion in 'Physics & Math' started by Vern, Aug 31, 2006.

1. DaleSpamTANSTAAFLRegistered Senior Member

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:bugeye: That is the definition of monochromatic.

-Dale

3. VernRegistered Senior Member

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One photon does not produce an infinitely small piece of a spectrum. One photon consists of one wave length. So if monochromatic means an infinitely small piece of a spectrum, it is not related to photon acton.

Last edited: Sep 2, 2006

5. DaleSpamTANSTAAFLRegistered Senior Member

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Vern, talking with you is confusing. These two sentences are compatible.
But having said that what do you mean by this sentence:
Apparently you don't mean monochromatic, so I assume that you are talking about the physical size of a photon in some sense. Of course, since it is not monchromatic there is not a single wavelength that you could attribute to the photon any more than there is a single frequency, so this notion of photon size is inherently a little ambiguous.

-Dale

7. 2inquisitiveThe Devil is in the detailsRegistered Senior Member

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DaleSpam, we know what you are protecting. The assumption of all relativity theories that it makes no difference which frame is considered stationary and which frame is in motion, the assumption that the observer doing the measuring is always in the 'rest frame'. When an electron jumps orbit, a photon of a descrete energy, momentum and wavelength is emitted. A measuring device with a relative velocity wrt the emitter will record different energies, momentums and wavelengths for the emitted photon. By assuming the measuring device is 'at rest', that would indicate the photon was emitted at a different energy, momentum and wavelength from a moving emitter. Relativity thoeries would fail if the emitter were given a 'preferred' status as the rest frame.

8. DaleSpamTANSTAAFLRegistered Senior Member

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First, this has nothing whatsoever to do with the present discussion I am having with Vern. SR would be just fine with monochromatic photons.

Second, according to SR you can use any inertial frame, regardless if the emitter, the detector, or neither are at rest. I already went through the math about this point on one of our Doppler threads. Instead of spouting unsupported nonsense why don't you do the math yourself this time, it might help you learn something useful. Obviously my demonstration last time didn't help. Try using SR to derive the relativistic Doppler effect. Try it from a frame where the emitter is stationary, from a frame where the detector is stationary, and from a frame where they are both moving. You will learn alot if you are willing to put forth some effort.

-Dale

9. Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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To Vern:

My ISP has not functioned for most of the day so I am writing off line to post when it returns. Even if you have clarified all your verbal terms, I think it still useful to build a mathematical model so we can be more precise in stating and understanding you ideas

Thus, I suggest we reduce the universe to just two parallel-traveling identical photons, A and B and concern our selves only with the first order effect of B upon A. By “first order” I mean that we consider the effect or changes in A caused by the existence of B but do not recognize any changes in B that these changes in A would then cause upon B.

Photon A is traveling in the positive x direction exactly along the x-axis. of a Cartesian coordinate system. Both have their electric fields perpendicular to the xy plane. Lets assume that the photons are separated by distance “D.” Thus, we can assume photon B is traveling along the line y = D.

Furthermore, lets model both of them (really their “unity maximum” electric field strength) at t=0 as of the form:

E = sin(kx -wt), where k is 2pi/ lambda, the "wavelength" (of this sine function, not the photon, as there is no single wavelength associated with any finite length photon.)
for x>0 and x< lambda E9.
Note E = 0 elsewhere at t =0.

{I admit that regions near the x-axis, in the range on the x-axis where the sine wave is, there may be some field for any real photon. I.e. photons, are not “infinitely skinny“, I don’t think. Anyway, lets assume this model, at least to first order.
I do not know the mechanism by which photon B has influence upon photon A, but whatever it is, if it exists at all, (and I still have strong doubts that it does), I do not mean to exclude this mechanism by my statement that E = 0 elsewhere. I just think, that at least initially, we can be somewhat vague on the mechanism, any electric fields associated with it (your “electric gravity,” - I assume) or whatever it is you are trying to show. After I can understand you better, perhaps we will need to return to this question more carefully.}

That is, both photons are E9 cycles of a sin wave long. Lets assume this shape and length is constant, at least to first order, (no dispersion etc.). Thus, at some later time, when both photons have traveled 100 lambda, the region x < 100 lambda will be, to first order at least, field free (E=0)

Now I can precisely ask (and hopefully you can answer) some questions:

(1) At time t = 0, is the “center of photon A” at (5E8,0,0)? If not, where? (Note here and later, lambda = 1. That is lambda is now also my unit of length.)

(2) Where, at t =0, is the “point in front” of A caused by the existence of B?

(3) How does the distance between “the center of A” and the “pointing front of A” depend upon the distance D? (I am assuming this distance is a constant and not a function of time. - Correct me if this is wrong.)

(4) How is “in front” distinguished from “behind”? Note that both photons have always been exactly side-by-side so that any (electric?) forces (with instanteous propagation speed) must act symmetrically. Stating this in alternative terms: from either photon’s POV the other is just static and always by its side, but D away.

I am betting you will want to assume that the interaction forces do not act instanteously. So that means by the time, t1, when the effect B is creating at t = 0 reaches the x-axis, presumably symmetric about the point (5E9,0,0) the center of A (if it was as I guess at that point when t =0 is now at ({5E8+d}, 0,0) where “d’ is some small number of wavelengths, perhaps 100 for some suitable value of D (or t1). Lets assume this “suitable” D (or t1) value.

Then

(5) It would seem the center of A at t1 is at ({5E8+100},0,0) and the point the the effect of B is symmetric about is (5E8,0,0) which to me implies that the net effect of B on A is BEHIND THE CENTER of A, not in front as you assert. Can you explain / answer my question (4) using this numerical examples values?

I could posses additional questions but this is enough for now. I am not trying to “shoot your theory down,” but to make you state it accurately and self consistently. Your previous “words only” version was frankly very confusing to me and seemed to inconsistently use your same words at different times to mean different things. I could not even be sure of the dimensionality of some terms (point is 0D, line is 1D etc. but same term for both? Etc?)

If you do not like my “truncated E9 cycle sine wave” model of the photon, please suggest a mathematical alternative. Mine is not an accurate model I am sure, perhaps there is no known accurate one, but it is simple, and I think it can at least serve to clarify what you are trying to state.

PS, now that I can connect again, I did so quickly without ready the recent posts - will read now if ISP stays up.

Last edited by a moderator: Sep 3, 2006
10. Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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To Vern (after reading recent posts. My ISP is still up):

Dale has been trying to clarify it for you (quite correctly I think). You said you had been an electrical engineer (or something like that). So I will assume you are familiar with Fourier analysis. Probably have at least once in your life you have been forced to calcualte the spectral components of various functions, such as a single finite width step with return to zero which I now try to draw by typing:

……………….......….....______
_______________|...........|________________

(ignore all dots. View only solid lines. Assume that flat horizontals extend forever.)

Thus you know that most any function can be resolved in to the sum of frequency components. For example this one, which I will call “A”

________/\/\/\/\/\/\/\/\_____________________

And this one I will call “B”

________/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\________

The frequency content of B is less wide than that of A.

Now if I wanted to make the frequency content of a function like A or B even more sharply defined, I would need to have many more of the /\/\/\ cycles, but even if they extended forever in both directions, I would not get a single monochromatic frequency because of the corners (at least not in the standard Fourier analysis with “orthogonal base functions” of sin and cosine and the harmonics of their fundamentals.)

Surely you, as old EE engineer, know all this.

Thus you must be postulating (and I do not think you are) that a photon extends across the entire universe (and beyond) so that it can have only a single frequency or energy, as you claim it “really does”

Now in my other post, I suggested a mathematical model of a photon which had E9 cycles of pure sine wave and then was identically zero amplitude before and after this oscillatory section. The frequency content of that photon is very well defined, but not exactly a single frequency (not monochromatic)

Even if you do not see how this related to the uncertainty principle or the fact that the transition probability of an atom in the excited state to the ground state which produces that well defined photon being very low or some other things, you must, as old EE engineer realize that no photon is strictly monochromatic. This is just a fact of reality, not a measurement difficulty.

Dale has been telling you the truth about reality. I hope putting it to you in these terms, which surely all old EEs must know, is persuasive.

Last edited by a moderator: Sep 3, 2006
11. URIIMURegistered Senior Member

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>> he final irreducible constituent of all physical reality was the electromagnetic field. >>

actually an aether of magnetism... with motion creating the electric vector

and the math is not so hard

There is an experiment where this assertion can be verified (for example, there are others)

It has all to do with the Lagrangian L1 and L2 points

Employing gravity as a one way force, leads to erroneous results for these positions.
IMO

But I intend to leave this.... sorry I will not expand.

12. VernRegistered Senior Member

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Hi Billy T; yes; I did understand what DaleSpam was trying to teach me. In my eingineering experiences we always assumed some degree of precision, and didn't usually talk in terms of absolutes. Our communication problem was, I think, that I was assuming a degree of precision.

I have thought about your model of two parallel photons and the difficulty of having one as a gravitational source for the other. This can't happen if the gravitational communication is limited to the speed of light.

In the gravitational model I'm trying to describe, one photon whould have to trail the other so that it is within its expanding fields. I had thought before I attended the 1995 NYAC conference titled, "Fundamental Problems in Quantum Theory", that two photons traveling parallel could not attract each other bacause of the speed-of-light limitation. However, at that conference, there were many papers presented on entangled photons which seemed to communicate with each other instantly.

So now I leave that open. I'll study your post some more Billy T, and see if I can comment on each issue. I really appriciate you taking the time to do this.

Last edited: Sep 3, 2006
13. VernRegistered Senior Member

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I think I finally figured out what you are saying. I know that nothing is ever absolute. Maybe I should say that to some degree of precision a photon consists of one wavelength. If you want it to be 1.00000000000fuzzy stuff. that's Ok.

14. Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Thanks for that at least.

15. Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I hope you understand this has to due with the reality, or lack of existence of monochromatic photons, and nothing to due with limits of measurements.

In some ways, the finite length of the photon (see discussion in my post with typed graphs) is even stronger than the uncertainity principle, which does set fundamental limit on the measurement results, (unless your measurement takes infinite time to complete)

16. VernRegistered Senior Member

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Hi Billy T.

I'm responding here off-line as I draw this model, I'm thinking of the x axis being left-right, the y up-down, and the z in the direction of travel.

In this schematic, the y axis represents the amplitude of the electric field at times T1 through T5 and the x axis represents the amplitude of the magnetic field. I call this the classic photon model because it is the way it was presented to me in school around 50 years ago.

The first addition I make to this classic model is the idea of saturation of points in space that exist along the z axis. I assume some finite size for the points, like a granular space, but this is not critical to the notion. Edit: On second thought it is critical if we try and calclulate the values, to avoid a singularity. Maybe a Planck's Point would work. BTW, this electrical saturation constant must also be the seat of the charge of the electron.

By saturation, I mean that the electric and magnetic amplitude at the moving points T2 and T4, are the maximum value that electric and magnetic fields can possibly reach in space. This is an important characteristic because it is the seat of "electric gravity" as you have coined it

I don't know the values of the Electric and Magnetic saturation amplitudes, but with your math capability you might be able to back into them by starting with Planck's constant, which must derive from this photon characteric.

I'll go ahead and post this because we have to nail down this idea of saturation before anything else will make sense.

Last edited: Sep 3, 2006
17. DaleSpamTANSTAAFLRegistered Senior Member

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So what do you mean by "a photon consists of one wavelength"? Are you talking about spectral content? Are you describing its spatial size? Is a photon time limited and band unlimited? Are you talking about the principal wavelength or some sort of average wavelength? Are you trying to describe a photon purely in terms of observable quantities or are you trying to assume some underlying non-observable nature? What are the various quantities in your theory, how do they relate to each other, and what predictions does your theory make?

I am not trying to nit-pick here, right now I am just trying to figure out what you are talking about. The main problem with this and most "alternative theories" is that either the author doesn't have a clear idea of how his own theory works and/or he doesn't communicate it clearly. This is the main reason that you should develop a mathematical framework and specific definitions. Otherwise you are going to get the feeling that everyone is dismissing your ideas without considering them when the truth is that you simply haven't given them any clear ideas to dismiss.

-Dale

18. DaleSpamTANSTAAFLRegistered Senior Member

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Are you taking this to be a literal and complete schematic of a single photon? Are you saying that in space a single photon lies along the z axis and occupies the region from T1 to T5? That the photon is attached to or somehow contains some mutually perpendicular but otherwise straight E and B fields?

If so, you have much worse problems than QM considerations. And you really misunderstood what the teachers were trying to tell you 50 years ago.

-Dale

19. VernRegistered Senior Member

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Hi DaleSpam; the orign of this speculation was that I was testing an old idea that prevailed around the turn of the twentieth century. That idea was that the final irreducible constituent of all physical reality was the electromagnetic field. That idea obviously is not compatible with QED, and most else of QM theory. So this whole concept lies outside of QM theory.

If an when it develops into something useful we can reconcile it with QM observations, not necessarily QM theory. You can't just say my theory don't agree with your theory therefore your theory is wrong.

Don't get me wrong, I don't have anything against QM theory; I find it very interesting; but to think about a universe in which nothing exists except electromagnetic fields, you have to crawl out of the QM box for awhile.

20. DaleSpamTANSTAAFLRegistered Senior Member

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Certainly not, I can say that your theory is unclear (and therefore useless) without reference to any other theory. Either your ideas are not clearly formulated in your own mind or you are having a hard time communicating them clearly.

Either way, you desperately need some clear definitions and some explicit math to go with your theory.

-Dale

21. VernRegistered Senior Member

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No; not literal nor complete; just a simple schematic to help get the idea across. I'm saying that a single photon moves through space. The z axis represents that path in space. The E and B fields radiate outward at the speed of light forever diminishing in amplitude as the inverse square of distance.

22. DaleSpamTANSTAAFLRegistered Senior Member

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This is getting worse by the minute. The E and B fields radiate outward from a photon? I thought the photon itself was the radiation. Are you saying that one photon will give birth to countless other photons as it travels along.

Get your theory together. It is useless talking like this.

-Dale

23. VernRegistered Senior Member

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I hesitate to call it a theory; I agree with you; I'm just having fun trying to prove that it's impossible for the universe to be comprised only of electromagnetic fields. I'm not investing too much in it because I know it's not going anywhere.